Ex 6.4, 3 - ABC and DBC are two triangles on same base BC - Area of similar triangles

Ex 6.4, 3 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.4, 3 - Chapter 6 Class 10 Triangles - Part 3 Ex 6.4, 3 - Chapter 6 Class 10 Triangles - Part 4

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 3 In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (𝑎𝑟(𝐴𝐵𝐶))/(𝑎𝑟(𝐷𝐵𝐶)) = 𝐴𝑂/𝐷𝑂 Given: ∆ ABC and ∆ DBC Having common base BC To prove: (𝑎𝑟 ∆ 𝐴𝐵𝐶)/(𝑎𝑟 ∆ 𝐷𝐵𝐶)=𝐴𝑂/𝐷𝑂 Proof: Since AO and OD are not part of ∆ 𝐴𝐵𝐶 and ∆ 𝐷𝐵𝐶 , we cannot directly use the theorem We know that Area of triangle = 1/2×𝐵𝑎𝑠𝑒×𝐴𝑙𝑡𝑖𝑡𝑢𝑑𝑒 Lets draw altitude AE [ AE ⊥𝐵𝐶 ] Hence, ar ∆ ABC=1/2×𝐵𝐶×𝐴𝐸 Similarly, for Δ DBC Let us draw altitude DF (DF ⊥ BC) Hence, ar ∆ DBC=1/2×𝐵𝐶×𝐷𝐹 Now, taking ratio (𝑎𝑟 ∆𝐴𝐵𝐶)/(𝑎𝑟 ∆𝐷𝐵𝐶) = (1/2 × 𝐵𝐶 × 𝐴𝐸)/(1/2 × 𝐵𝐶 × 𝐷𝐹) (𝑎𝑟 ∆𝐴𝐵𝐶)/(𝑎𝑟 ∆𝐷𝐵𝐶) = 𝐴𝐸/𝐷𝐹 Now in ∆ AOE and Δ DOF ∠ AEO = ∠ DFO ∠ AOE = ∠ DOF So, by using AA similarity criterion ∆ AOE ~ ∆ DOF If two triangle are similar their , corresponding sides are in the same ratio So, 𝐴𝐸/𝐷𝐹=𝐴𝑂/𝐷𝑂 Putting (2) in (1) (𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐵𝐶)/(𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐷𝐵𝐶)=𝐴𝐸/𝐷𝐹 (𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐵𝐶)/(𝑎𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐷𝐵𝐶)=𝐴𝑂/𝐷𝑂 Hence proved

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.